Trigonometric ratios in a right triangle

Learning outcome

Identify the three sides relative to an acute angle, calculate all six trigonometric ratios, and recover missing lengths from a given ratio.

Concepts and assumptions

Angles in this lesson are measured in degrees. Use a non-degenerate right triangle ABC with ∠C = 90° and reference angle θ = ∠A, so 0° < θ < 90°. All side lengths are positive and use matching units.

AB is the hypotenuse H, opposite the right angle. Relative to θ, BC is the opposite side O and AC the adjacent leg J. The adjacent leg is not the hypotenuse, even though both touch A.

Define: sin θ = O/H; cos θ = J/H; tan θ = O/J. cosec θ = H/O; sec θ = H/J; cot θ = J/O.

Every denominator is positive under these assumptions, so all six ratios exist. Ratios have no length unit because matching units cancel. Right triangles with the same acute angle are similar: corresponding sides scale together, leaving each ratio unchanged. Thus ratios depend on the angle, not triangle size.

The opposite and adjacent legs exchange roles when the reference angle changes to B; the hypotenuse stays fixed. Pythagoras, O² + J² = H², supplies a missing side. Choose the positive square root because lengths are positive. Keep exact fractions and roots unless rounding is requested.

Worked examples

Example 1 — Six ratios. Let BC = 9 cm and AC = 12 cm. Find the ratios for A.

AB = √(9² + 12²) = √225 = 15 cm. Thus sin A = 9/15 = 3/5, cos A = 12/15 = 4/5, tan A = 9/12 = 3/4. Reversing these fractions gives cosec A = 5/3, sec A = 5/4, cot A = 4/3.

Example 2 — Recovering lengths. For an acute θ, sin θ = 5/13 and H = 39 m.

O = (5/13) × 39 = 15 m. J = √(39² − 15²) = √1296 = 36 m. Hence cos θ = 36/39 = 12/13 and tan θ = 15/36 = 5/12.

Example 3 — Changing the reference angle. Let AB = 25 cm and AC = 7 cm.

BC = √(25² − 7²) = √576 = 24 cm. For A, opposite:adjacent = 24:7, so tan A = 24/7. For B, these roles reverse: tan B = 7/24, sin B = 7/25, and cos B = 24/25.

Common mistakes

Choosing sides without naming the reference angle; confusing a leg with the hypotenuse; attaching length units to ratios; using negative side lengths.

Practice questions

1. In ABC with ∠C = 90°, BC = 20 cm and AC = 21 cm. Find AB, sin A, cos A and tan A.
2. For acute θ, cos θ = 8/17 and H = 34 cm. Find J, O and tan θ.
3. For acute θ, tan θ = 9/40. Find sin θ and sec θ.
4. In ABC with ∠C = 90°, BC = 12 cm and AC = 35 cm. Find cot A and cot B.

Worked solutions

1. AB = √(400 + 441) = 29 cm. Therefore sin A = 20/29, cos A = 21/29, tan A = 20/21.
2. J = 34 × 8/17 = 16 cm. O = √(1156 − 256) = 30 cm. Thus tan θ = 30/16 = 15/8.
3. Write O = 9k and J = 40k, with k > 0. H = √(81 + 1600)k = 41k. Hence sin θ = 9/41 and sec θ = 41/40.
4. Cotangent is adjacent/opposite. For A this is 35/12; for B the legs exchange roles, giving 12/35.